Question

What is the sum of $\angle BAD+\angle BPR+\angle BCD+\angle BQR$ in the above figure?

• A. ${540}^{\circ }$
• B. ${360}^{\circ }$
• C. ${240}^{\circ }$
• D. None of the above

Answer 1

The correct option is B ${360}^{\circ }$

In cyclic quadrilateral $BADC$,

$\angle BAD+\angle BCD={180}^0$.....[opposite angles of a cyclic quadrilateral are supplementary]

In cyclic quadrilateral $BPRQ$,

$\angle BPR+\angle BQR={180}^0$...[opposite angles of a cyclic quadrilateral are supplementary]

Hence the sum of

$\angle BAD+\angle BPR+\angle BCD+\angle BQR$ $={180}^o+{180}^o={360}^o$.

Therefore, option $B$ is correct.